Copyright 2000-2016 Networking Laboratory
Lecture 3.
Recurrences / Heapsort
T. H. Cormen, C. E. Leiserson and R. L. Rivest
Introduction to Algorithms, 3rd Edition, MIT Press, 2009
Sungkyunkwan University
Hyunseung Choo
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Overview for Recurrences
Define what a recurrence is
Discuss three methods of solving recurrences
Substitution method
Recursion-tree method
Master method
Examples of each method
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Definition
A recurrence is an equation or inequality that describes a
function in terms of its value on smaller inputs.
Example from MERGE-SORT
T(n) =(1) if n=1
2T(n/2) + (n) if n>1
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Technicalities
Normally, independent variables only assume integral
values
Example from MERGE-SORT revisited
T(n) =(1) if n=1
T(n/2) + T(n/2) + (n) if n>1
For simplicity, ignore floors and ceilings – often
insignificant
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Technicalities
Value of T(n) assumed to be small constant for small n
Boundary conditions (small n) are also glossed over
T(n) = 2T(n/2) + (n)
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Substitution Method
Involves two steps:
Guess the form of the solution
Use mathematical induction to find the constants and show
the solution works
Drawback
Applied only in cases where it is easy to guess at solution
Useful in estimating bounds on true solution even if
latter is unidentified
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Substitution Method
Example:
T(n) = 2T(n/2) + n
Guess:
T(n) = O(n lg n)
Prove by induction:
T(n) cn lg n
for suitable c>0.
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Inductive Proof
We’ll not worry about the basis case for the moment –
we’ll choose this as needed – clearly we have:
T(1) = (1) cn lg n
Inductive hypothesis: For values of n < k the inequality holds, i.e., T(n) cn lg n
We need to show that this holds for n = k as well.
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Inductive Proof
In particular, for n = k/2, the inductive hypothesis
should hold, i.e.,
T(k/2) c k/2 lg k/2
The recurrence gives us:
T(k) = 2T(k/2) + k
Substituting the inequality above yields:
T(k) 2[c k/2 lg k/2] + k
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Inductive Proof
Because of the non-decreasing nature of the functions
involved, we can drop the “floors” and obtain:
T(k) 2[c (k/2) lg (k/2)] + k
Which simplifies to:
T(k) ck (lg k lg 2) + k
Or, since lg 2 = 1, we have:
T(k) ck lg k ck + k = ck lg k + (1 c)k
So if c 1, T(k) ck lg k Q.E.D.
Algorithms
Practice Problems
Use the substitution method to show that
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)()( 2nOnT
1)1(
)3/(7)( 2
T
nnTnT
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Recursion-Tree Method
Straightforward technique of coming up with a good
guess
Can help the Substitution Method
Recursion tree: visual representation of recursive call
hierarchy where each node represents the cost of a
single subproblem
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Recursion-Tree Method
T(n) = 3T(n/4) + (n2)
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Recursion-Tree Method
T(n) = 3T(n/4) + (n2)
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Recursion-Tree Method
T(n) = 3T(n/4) + (n2)
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Recursion-Tree Method
T(n) = 3T(n/4) + (n2)
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Recursion-Tree Method
Gathering all the costs together:
T(n) = (3/16)icn2 + (nlog43) i=0
log4n1
T(n) (3/16)icn2 + o(n) i=0
T(n) (1/(13/16))cn2 + o(n)
T(n) (16/13)cn2 + o(n)
T(n) = O(n2)
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Recursion-Tree Method
T(n) = T(n/3) + T(2n/3) + O(n)
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Recursion-Tree Method
An overestimate of the total cost:
T(n) = cn + (nlog3/22) i=0
log3/2n1
T(n) = O(n lg n) + (n lg n)
Counter-indications:
T(n) = O(n lg n)
Notwithstanding this, use as “guess”:
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Substitution Method
Recurrence:
T(n) = T(n/3) + T(2n/3) + cn
Guess:
T(n) = O(n lg n)
Prove by induction:
T(n) dn lg n
for suitable d>0 (we already use c)
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Inductive Proof
Again, we’ll not worry about the basis case
Inductive hypothesis: For values of n < k the inequality holds, i.e., T(n) dn lg n
We need to show that this holds for n = k as well.
In particular, for n = k/3, and n = 2k/3, the inductive
hypothesis should hold…
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Inductive Proof
That is
T(k/3) d k/3 lg k/3
T(2k/3) d 2k/3 lg 2k/3
The recurrence gives us:
T(k) = T(k/3) + T(2k/3) + ck
Substituting the inequalities above yields:
T(k) [d (k/3) lg (k/3)] + [d (2k/3) lg (2k/3)] + ck
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Inductive Proof
Expanding, we get
T(k) [d (k/3) lg k d (k/3) lg 3] +
[d (2k/3) lg k d (2k/3) lg(3/2)] + ck
Rearranging, we get:
T(k) dk lg k d[(k/3) lg 3 + (2k/3) lg(3/2)] + ck
T(k) dk lg k dk[lg 3 2/3] + ck
When dc/(lg3 (2/3)), we should have
the desired:
T(k) dk lg k
Algorithms
Practice Problems
Use the recursion tree method to show that
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)()( 2nnT 2)2/()4/()( nnTnTnT
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Master Method
Provides a “cookbook” method for solving recurrences
Recurrence must be of the form:
T(n) = aT(n/b) + f(n)
where a1 and b>1 are constants and f(n) is an
asymptotically positive function.
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Master Method
Theorem 4.1:
Given the recurrence previously defined, we have:
1. If f(n) = O(n logba)
for some constant >0,
then T(n) = (n logba)
2. If f(n) = (n logba),
then T(n) = (nlogba lg n)
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Master Method
3. If f(n) = (n logba+)
for some constant >0,
and if
af(n/b) cf(n)
for some constant c<1
and all sufficiently large n,
then T(n) = (f(n))
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Example
Estimate bounds on the following recurrence:
Use the recursion tree method to arrive at a “guess” then verify
using induction
Point out which case in the Master Method this falls in
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Recursion Tree
Recurrence produces the following tree:
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Cost Summation
Collecting the level-by-level costs:
A geometric series with base less than one;
converges to a finite sum, hence, T(n) = (n2)
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Exact Calculation
If an exact solution is preferred:
Using the formula for a partial geometric series:
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Exact Calculation
Solving further:
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Master Theorem (Simplified)
Algorithms
Practice Problems
Use Master theorem to find asymptotic bound.
a.
b.
c.
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nnTnT )2/(4)(
2)2/(4)( nnTnT
3)2/(4)( nnTnT
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Introduction for Heapsort
Heapsort
Running time: O(n lg n)
Like merge sort
Sorts in place: only a constant number of array elements
are stored outside the input array at any time
Like insertion sort
Heap
A data structure used by Heapsort to manage information
during the execution of the algorithm
Can be used as an efficient priority queue
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Perfect Binary Tree
For binary tree with height h
All nodes at levels h–1 or less have 2 children (full)
h = 1 h = 3h = 2
h = 0
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Complete Binary Trees
For binary tree with height h
All nodes at levels h–2 or less have 2 children (full)
All leaves on level h are as far left as possible
h = 1
h = 2
h = 0
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Complete Binary Trees
h = 3
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Heaps
Two key properties
Complete binary tree
Value at node
Smaller than or equal to values in subtrees
Greater than or equal to values in subtrees
Example max-heap
Y X
Z X
Y
X
Z
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Heap and Non-heap Examples
Min-heaps Non-heaps
6
2
22
8 45 25
6
2
22
8 45 25
8
6 455
6 22
25
5
5 45
5
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Binary Heap
An array object that can be viewed as a nearly
complete binary tree
Each tree node corresponds to an array element
that stores the value in the tree node
The tree is completely filled on all levels except possibly the lowest,
which is filled from the left up to a point
A has two attributes
length[A]: number of elements in the array
heap-size[A]: number of elements in the heap stored within A
heap-size[A] length[A]
max-heap and min-heap
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A Max-heap
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Length and Heap-Size
11 7
711
Length = 10
Heap-Size = 7
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Heap Computation
Given the index i of a node, the indices of its parent,
left child, and right child can be computed simply:
12:)(
2:)(
2/:)(
ireturniRIGHT
ireturniLEFT
ireturniPARENT
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Heap Computation
16
14
8 7
142
10
9 3
0
1
2
3
parent(i) = floor(i/2)
left-child(i) = 2i
right-child(i)= 2i +1
1
2 3
4 5 6 7
8 9 10
16 14 10 8 7 9 3 2 4 1
1 2 3 4 5 6 7 8 9 10
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Heap Property
Heap property
The property that the values in the node must satisfy
Max-heap property, for every node i other than the root
A[PARENT(i)] A[i]
The value of a node is at most the value of its parent
The largest element in a max-heap is stored at the root
The subtree rooted at a node contains values
on larger than that contained at the node itself
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Heap Height
The height of a node in a heap
The number of edges on the longest simple downward path from
the node to a leaf
The height of a heap is the height of its root
The height of a heap of n elements is (lg n)
Exercise 6.1-2 on page 129
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Heap Procedures
MAX-HEAPIFY
Maintains the max-heap property
O(lg n)
BUILD-MAX-HEAP
Produces a max-heap from an unordered input array
O(n)
HEAPSORT
Sorts an array in place
O(n lg n)
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Maintaining the Heap Property
MAX-HEAPIFY
Inputs: an array A and an index i into the array
Assume the binary tree rooted at LEFT(i) and RIGHT(i) are max-heaps,
but A[i] may be smaller than its children
violate the max-heap property
MAX-HEAPIFY let the value at A[i] floats down in the max-heap
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Example of MAX-HEAPIFY
16
4
14 7
18
10
9 3
4 < 14
2
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Example of MAX-HEAPIFY
16
14
4 7
18
10
9 3
4 < 8
2
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Example of MAX-HEAPIFY
16
14
8 7
14
10
9 3
2
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MAX-HEAPIFY
Extract the indices of LEFT and RIGHT
children of i
Choose the largest of A[i], A[l], A[r]
Float down A[i] recursively
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MAX-HEAPIFY
Recursive versionInteractive version
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Running Time of MAX-HEAPIFY
(1) to find out the largest among
A[i], A[LEFT(i)], and A[RIGHT(i)]
Plus the time to run MAX-HEAPIFY on a
subtree rooted at one of the children of node i
The children’s subtrees each have size
at most 2n/3 (why?)
the worst case occurs when the last row of the tree is
exactly half full
T(n) T(2n/3) + (1)
By case 2 of the master theorem
T(n) = O(lg n)
7/11 = 0.63
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Heapify Example
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Building a Max-Heap
Observation: A[(n/2+1)..n] are all leaves of the tree
Exercise 6.1-7 on page 130
Each is a 1-element heap to begin with
Upper bound on the running time
O(lg n) for each call to MAX-HEAPIFY, and call n times O(n lg n)
Not tight
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Building a
Max-Heap
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Loop Invariant
At the start of each iteration of the for loop of lines 2-3, each node i+1,
i+2, .., n is the root of a max-heap
Initialization: Prior to the first iteration of the loop, i = n/2. Each
node n/2+1, n/2+2,.., n is a leaf and the root of a trivial max-heap.
Maintenance: Observe that the children of node i are numbered
higher than i. By the loop invariant, therefore, they are both roots of
max-heaps. This is precisely the condition required for the call
MAX-HEAPIFY(A, i) to make node i a max-heap root. Moreover, the
MAX-HEAPIFY call preserves the property that nodes i+1, i+2, …, n
are all roots of max-heaps. Decrementing i in the for loop update
reestablishes the loop invariant for the next iteration.
Termination: At termination, i=0. By the loop invariant, each node 1, 2, …, n
is the root of a max-heap. In particular, node 1 is.
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Cost for Build-MAX-HEAP
Heap-properties of an n-element heap
Height = lg n
At most n/2h+1 nodes of any height h
Exercise 6.3-3 on page 135
)()2
()2
()( 2 0
lg
0
lg
01
nOh
nOh
nOhOn
hh
n
hh
n
hh
Ignore the constant ½ 2
)2
11(
21
2 20
hh
h
20 )1( x
xkx
k
k
(for |x| < 1)
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Heapsort
Using BUILD-MAX-HEAP to build a max-heap on the input array A[1..n],
where n=length[A]
Put the maximum element, A[1], to A[n]
Then discard node n from the heap by decrementing heap-size(A)
A[2..n-1] remain max-heaps, but A[1] may violate
call MAX-HEAPIFY(A, 1) to restore the max-heap property
for A[1..n-1]
Repeat the above process from n down to 2
Cost: O(n lg n)
BUILD-MAX-HEAP: O(n)
Each of the n-1 calls to MAX-HEAPIFY takes time O(lg n)
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Example: Heapsort
16
14
8 7
142
10
9 3
1
2 3
4 5 6 7
8 9 10
16 14 10 8 7 9 3 2 4 1
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Example: Heapsort (2)
14
8
4 7
1612
10
9 3
1
2 3
4 5 6 7
8 9 10
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Example: Heapsort (3)
10
8
4 7
16142
9
1 3
1
2 3
4 5 6 7
8 9 10
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Example: Heapsort (4)
9
8
4 7
161410
3
1 2
1
2 3
4 5 6 7
8 9 10
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Example: Heapsort (5)
7
4
1 2
161410
3
8 9
1
2 3
4 5 6 7
8 9 10
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Example: Heapsort (6)
4
2
1 7
161410
3
8 9
1
2 3
4 5 6 7
8 9 10
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Example: Heapsort (7)
1
2
4 7
161410
3
8 9
1
2 3
4 5 6 7
8 9 10
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Example: Heapsort (8)
1
2
4 7
161410
3
8 9
1
2 3
4 5 6 7
8 9 10
1 2 3 4 7 8 9 10 14 16
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Heapsort Algorithm
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Heapsort Algorithm
Algorithms
Heap & Heap Sort Algorithm
Video Content
An illustration of Heap and Heap Sort.
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Algorithms
Heap & Heap Sort Algorithm
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Priority Queues
We can implement the priority queue ADT with a heap. The operations are:
Max(S) – returns the maximum element
Extract-Max(S) – remove and return the maximum element
Insert(x,S) – insert element x into S
Increase-Key(S,x,k) – increase x’s value to k
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Extract-Max
Heap-Maximum(A) return A[1]
Heap-Extract-Max(A)
1. if heapsize[A] < 1
2. then error “heap underflow”
3. max A[1]
4. A[1] A[heapsize[A]]
5. heapsize[A] heapsize[A] –1
6. Max-Heapify(A,1)
7. return max
(1)
O(lg n)
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Increase-Key
Heap-Increase-key(A, i, key)
1. if key < A[i]
2. then error “new key smaller than existing one”
3. A[i] key
4. while i > 1 and A[parent(i)] < A[i]
5. do Exchange(A[i], parent(A[i]))
6. i parent(i)
O(lg n)
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Example: increase key (1)
16
14
8 7
142
10
9 3
1
2 3
4 5 6 7
8 9 10
increase 4 to 15
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Example: increase key (2)
16
14
8 7
1152
10
9 3
1
2 3
4 5 6 7
8 9 10
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Example: increase key (3)
16
14
15 7
182
10
9 3
1
2 3
4 5 6 7
8 9 10
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Example: increase key (4)
16
15
14 7
182
10
9 3
1
2 3
4 5 6 7
8 9 10
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Insert-Max
Heap-Insert-Max(A, key)
1. heapsize[A] heapsize[A] + 1
2. A[heapsize[A]] -∞3. Heap-Increase-Key(A, heapsize[A], key)
O(lg n)
Algorithms
Practice Problems
Show the resulting heap after insert 20 into the following
heap
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